I realized the only example of a nonmeasurable set I had seen (a vitali set) crucially relies on the translation invariance of the lebesgue measure. I looked around for other examples of nonmeasurable sets, and struggled to find any. This led me to ask if there are measures which can be defined on the entire powerset algebra.
A moment's thought shows the dirac measure works, and indeed you can take the measure $\sum_{r_i \in \mathbb{Q}} 2^{-i} \delta_{r_i}$ to make this example slightly less trivial... But what about nonatomic measures?
I'm casually acquainted with set theory, and I thought this might depend on your choice of foundations. So I asked a set theorist friend of mine, and he pointed me towards "real-valued measurable cardinals". The wikipedia page says
"A real valued measurable cardinal less than or equal to $\mathfrak{c}$ exists if and only if there is a countably additive extension of the lebesgue measure to all sets of real numbers if and only if there is an atomless probability measure on the power set of some nonempty set"
Where can I find a reference for this equivalence? Or, if it isn't hard, can someone post a proof here? I am comfortable with Set Theory at the level of Kunen's "easy consistency proofs". In particular, while I know the idea of forcing, I've never worked with it in practice.
Thanks in advance ^_^
This does not answer the explicit questions but the first implicit one above.
Other important examples of nonmeasurable sets are Bernstein's. Those are based on the inner regularity of measure: Every non null set includes a non null (and, a fortiori uncountable) closed set.
A Bernstein set $B\subset \mathbb{R}$ satisfies that $F\nsubseteq B, B^c$ for every uncountable closed set $F$. They can not be null (consider $B^c$) but can't have positive measure (now consider $B$). Such a $B$ can be constructed by transfinite recursion, enumerating the all uncountable closed subsets $\{F_\alpha\}_\alpha$ in order type $2^{\aleph_0}$, and choosing different $b_\alpha,c_\alpha\in F_\alpha \setminus\{b_\beta,c_\beta: \beta<\alpha\}$. (For this to work, you must recall the Cantor-Bendixson Theorem and that perfect subsets of $\mathbb{R}$ have cardinality $2^{\aleph_0}$.) Hence $B:=\{b_\beta: \beta<2^{\aleph_0}\}$ is a Bernstein set.
Bernstein sets are also important since they are not determined.